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Rus_ich [418]
3 years ago
13

In space, astronauts don’t have gravity to keep them in place. That makes doing even simple tasks difficult. Gene Cernan was the

first astronaut who worked on a task outside a spaceship. He said of the experience, “Every time I’d push or turn a valve, it would turn my entire body at zero gravity. I had nothing to hold on to.” As he worked, Gene Cernan’s heart rate and temperature went so high that his fellow astronauts worried that he wouldn’t survive.
Think about routine tasks that astronauts might need to do inside and outside a spaceship. Choose several tasks, and describe the features the ship and spacesuits should have to account for zero gravity as the astronaut completes the task. Use Newton’s laws of motion in your analysis.
Physics
1 answer:
PIT_PIT [208]3 years ago
8 0

going to the toilet, and keeping cleant. no nasty bits floating around but they are weightless !

eating. pushing a fork would push back by newton 3. can't brace themselves to cut

You might be interested in
How do I calculate the tension in the horizontal string?
matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

T_1\cos 30\degree-m\cdot g=0

Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0
1 year ago
Which of the following is most needed for cosmotologists to study the age of the universe
Hatshy [7]
If its not Distance traveled then its energy
5 0
3 years ago
A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

4 0
2 years ago
The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter
GenaCL600 [577]

Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

Q= required charge

Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

diameter of the conductor is 4.20 mm

But Radius= diameter/2= 4.20/2=

The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then

radius of the conductor is 0.0021m.

We can now calculate the area of the conductor which is

A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

We can proceed to calculate the current density below

J = 140/13.85*10^-6 = 10108303A/m

According to the listed reference:

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s

Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

7 0
2 years ago
Mary was looking out of the window she saw lightening and then heard thunder a few seconds later explain why she saw lightening
alekssr [168]

Explanation:

It is based upon the fact that " The light travels faster then sound." As the speed of light is faster then the speed of sound, light travels 300,000 km per second and sound travels 1192 km per hour. That is why we observe the lightening first and hear the the sound of thunder later.

        You can do this experiment by yourself. Once you see the lightening start counting the seconds until you hear the sound of thunder.Then divide the seconds by 5, you will find out how many miles away the lightening strike was.

3 0
3 years ago
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